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(D) According to the de Broglie equation,the wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$.Substituting the given values:$\lambda = \frac{

Vedclass · Accepted Answer

$0.40$

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(D) According to the de Broglie equation,the wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$.Substituting the given values:$\lambda = \frac{6.63 	imes 10^{-34} \, J s}{(1.67 	imes 10^{-27} \, kg) 	imes (1.0 	imes 10^3 \, m s^{-1})}$$\lambda = \frac{6.63}{1.67} 	imes 10^{-34 + 27 - 3} \, m$$\lambda \approx 3.97 	imes 10^{-10} \, m$To convert meters to nanometers $(1 \, nm = 10^{-9} \, m)$:$\lambda = 3.97 	imes 10^{-10} \, m = 0.397 	imes 10^{-9} \, m \approx 0.40 \, nm$.

Calculate the wavelength in nanometers associated with a particle moving with a velocity of $1.0 \times 10^3 \, m s^{-1}$. (Given: mass $m = 1.67 \times 10^{-27} \, kg$ and Planck's constant $h = 6.63 \times 10^{-34} \, J s$)

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